solved exercise on congruent triangles

EXERCISE

1. In figure 5-8 Triangle ABC is an isosceles triangle with AB=AC. If A  80o , then B is a. 100o b. 80o c. 50o d. 130o Sol: Correct option is (c)



In isosceles Triangle ABC , since AB=AC

Therefore B  C (Angles opposite equal sides of a triangle are equal)



Using Angle Sum Property of Triangle we have

A  B  C  180o



or 80o  B  B  180o



( Since B  C )



or 80o  2B  180o or 2B  180o  80o or 2B  100o or B  100o  50o 2



2. The measure of an exterior angle of an equilateral triangle is a. 120o b. 60o c. 90o d. 100o Sol: Correct option is (b)



We know that in an equilateral triangle, measure of all the angles is equal.

Therefore A  B  C



Using Angle Sum Property of Triangles we have

A  B  C  180o



or C  C  C  180o or 3C  180o or C  180o  60o 3



( Since A  B  C )



Therefore A  B  C  60o Since ACB and ACD form a linear pair Therefore ACB  ACD  180o or 60o  ACD  180o or ACD  180o  60o  120o 3. In figure, Triangle ABC is an isosceles triangle with AB=AC and exterior angle ACD=112o . Then the measure of A is a. 56o b. 112o c. 90o d. 44o Sol: Correct option is (d)



In Triangle ABC , since AB=AC

Therefore B  C (Equal sides of triangles have equal angles opposite them)



Now C  112o  180o or C  180o  112o  68o



(Linear pair)



Using Angle Sum Property of Triangles, we have

A  B  C  180o



or A  68o  68o  180o or A  136o  180o or A  180o  136o  44o



( Since B  C )



4. In figure 5-11, AB=AC, DB=DC, A  40o and D  100o . The measure of ACD is a. 60o b. 20o c. 30o d. 40o



Sol:



Correct option is (c)



In Triangle ABC , since AB=AC

Therefore B  C (Angles opposite equal sides of a triangle are equal)



Now, using Angle Sum Property of Triangles, we have

A  B  C  180o



or 40o  C  C  180o or 40o  2C  180o or 2C  180o  40o or 2C  140o or C  Now in 140o  70o 2

Triangle DBC , DB  DC Therefore DBC  DCB



… (1) (Angles opposite equal side are equal) (Angle Sum Property of Triangle) ( Since DBC  DCB )



D  DBC  DCB  180o



or 100o  DCB  DCB  180o or 100o  2DCB  180o



or 2DCB  180o  100o or 2DCB  80o or DCB  80o  40o 2



Since ACD and DCB are adjacent angles

Therefore ACD  DCB  ACB



or ACD  40o  70o or ACD  70o  40o  30o



( ACB  C  70o )



5. In figure 5-12, BAC  80o and AB=AC=CD. Then the measure of CAD is a. 10o b. 25o c. 50o d. 35o Sol: Correct option is (b)



In Triangle ABC , since AB=AC

Therefore B  C (angles opposite equal sides are equal)



Now, A  B  C  180o (Angle Sum Property of Triangle) or 80o  C  C  180o or 80o  2C  180o or 2C  180o  80o or 2C  100o 100o or C   50o 2 In Triangle ACD , since AC=CD

Therefore CAD  CDA



( Since B  C )



(angles opposite equal sides of triangles are equal)



Now C is the exterior angles of Triangle ACD

Therefore C  CDA  CAD (Exterior angle property)



or 50o  CAD  CAD or 50o  2CAD or 2CAD  50o or CAD  50o  25o 2



( Since CAD  CDA )